Страница 120 номер 415, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Заполнить таблицу (414—415). \[\begin{array}{|l|c|c|c|c|c|c|} \hline \text{Угол, }{}^\circ & 30 & & & & & \\ \hline \text{Угол, рад} & & \frac{\pi}{5} & & & 2 & \\ \hline \text{Радиус, см} & 2 & & 10 & 5 & & \\ \hline \text{Длина дуги, см} & & 2 & 5 & & & 10 \\ \hline \text{Площадь сектора, см}^2 & & & & 50 & 25 & 50 \\ \hline \end{array}\]
\[\begin{array}{|c|c|c|c|c|c|c|} \hline \text{Угол, }^{\circ} & 30 & 36 & \frac{90}{\pi} & \frac{720}{\pi} & \frac{360}{\pi} & \frac{180}{\pi} \\ \hline \text{Угол, рад} & \frac{\pi}{6} & \frac{\pi}{5} & 0{,}5 & 4 & 2 & 1 \\ \hline \text{Радиус, см} & 2 & \frac{10}{\pi} & 10 & 5 & 5 & 10 \\ \hline \text{Длина дуги, см} & \frac{\pi}{3} & 2 & 5 & 20 & 10 & 10 \\ \hline \text{Площадь сектора, см}^{2} & \frac{\pi}{3} & \frac{10}{\pi} & 25 & 50 & 25 & 50 \\ \hline \end{array}\]
1) \(30^{\circ} = \frac{30\pi}{180} = \frac{\pi}{6}, l = \alpha R = \frac{\pi}{6} \cdot 2 = \frac{\pi}{3}, S = \frac{R^{2}}{2}\alpha = 2 \cdot \frac{\pi}{6} = \frac{\pi}{3};\)
2) \(\frac{\pi}{5} = \left(\frac{180}{\pi} \cdot \frac{\pi}{5}\right)^{\circ} = 36^{\circ}, R = \frac{l}{\alpha} = 2 \cdot \frac{5}{\pi} = \frac{10}{\pi}, S = \frac{R^{2}}{2}\alpha = \frac{50}{\pi^{2}} \cdot \frac{\pi}{5} = \frac{10}{\pi};\)
3) \(\alpha = \frac{l}{R} = \frac{5}{10} = 0{,}5, 0{,}5 = \left(\frac{180}{\pi} \cdot 0{,}5\right)^{\circ} = \left(\frac{90}{\pi}\right)^{\circ};\)
\(S = \frac{R^{2}}{2}\alpha = 50 \cdot 0{,}5 = 25;\)
4) \(\alpha = \frac{2S}{R^{2}} = \frac{100}{25} = 4, 4 = \left(\frac{180}{\pi} \cdot 4\right)^{\circ} = \left(\frac{720}{\pi}\right)^{\circ}, l = \alpha R = 4 \cdot 5 = 20;\)
5) \(2 = \left(\frac{180}{\pi} \cdot 2\right)^{\circ} = \left(\frac{360}{\pi}\right)^{\circ}, R = \sqrt{\frac{2S}{\alpha}} = \sqrt{25} = 5, l = \alpha R = 5 \cdot 2 = 10;\)
6) \(S = \frac{R^{2}}{2} \cdot \frac{l}{R} = \frac{lR}{2} = 50, R = \frac{100}{10} = 10, \alpha = \frac{l}{R} = \frac{10}{10} = 1, 1 = \left(\frac{180}{\pi}\right)^{\circ}.\)
